Chapter 4 Trigonometry and the Unit Circle · 2018. 2. 4. · arc length central angle =...

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MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 1 of 85 Chapter 4 Trigonometry and the Unit Circle Section 4.1 Angles and Angle Measure Section 4.1 Page 175 Question 1 a) –4π is a clockwise rotation b) 750° is a counterclockwise rotation c) –38.7° is a clockwise rotation d) 1 radian is a counterclockwise rotation Section 4.1 Page 175 Question 2 a) 30° π 30 180 π 6 b) 45° π 45 180 π 4 c) –330° π 330 180 11π 6 d) 520° π 520 180 26π 9

Transcript of Chapter 4 Trigonometry and the Unit Circle · 2018. 2. 4. · arc length central angle =...

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 1 of 85

    Chapter 4 Trigonometry and the Unit Circle Section 4.1 Angles and Angle Measure Section 4.1 Page 175 Question 1 a) –4π is a clockwise rotation b) 750° is a counterclockwise rotation c) –38.7° is a clockwise rotation d) 1 radian is a counterclockwise rotation Section 4.1 Page 175 Question 2

    a) 30° π30180

    π 6

    b) 45° π45180

    π 4

    c) –330° π330180

    11π 6

    d) 520° π520180

    26π 9

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 2 of 85

    e) 90° π90180

    π 2

    f) 21° π21180

    7π 60

    Section 4.1 Page 175 Question 3

    a) 60° π60180

    π 31.05

    b) 150° π150180

    5π 6

    2.62

    c) –270° π270180

    3π 2

    4.71

    d) 72° π72180

    2π 5

    1.26

    e) –14.8° π14.8

    180148π180037π4500.26

    f) 540° π540180

    3π 9.42

    Section 4.1 Page 175 Question 4

    a) π 1806 6

    30

    b) 2π 2(180 )3 3

    120

    c) 3π 3(180 )8 8

    67.5

    d) 5π 5(180 )2 2

    450

    e) 1801 1π

    180π

    57.3

    f) 1802.75 2.75π

    495π

    157.6

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 3 of 85

    Section 4.1 Page 175 Question 5

    a) 2π 2(180 )7 7

    3607

    51.429

    b) 7π 7(180 )13 13

    126013

    96.923

    c) 2 2 1803 3 π

    120π

    38.197

    d) 1803.66 3.66π

    658.8π

    209.703

    e) 1806.14 6.14π

    1105.2π

    351.796

    f) 18020 20π

    3600π

    1145.916

    Section 4.1 Page 175 Question 6 a) An angle that measures 1 radian is in quadrant I.

    b) An angle that measures 225° is in quadrant II.

    c) An angle that measures 17π6

    is in

    quadrant II.

    d) An angle that measures 650° is in quadrant IV.

    e) An angle that measures 2π3

    is in

    quadrant III.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 4 of 85

    f) An angle that measures –42° is in quadrant IV.

    Section 4.1 Page 176 Question 7 a) 72° + 360° = 432° 72° – 360° = –288° For an angle of 72°, one positive coterminal angle is 432° and one negative coterminal angle is –288°.

    b) 3π 11π2π4 4 3π 5π 2π

    4 4

    For an angle of 3π4

    , one positive coterminal angle is 11π4

    and one negative coterminal

    angle is – 5π4

    .

    c) –120° + 360° = 240° –120° – 360° = –480° For an angle of –120°, one positive coterminal angle is 240° and one negative coterminal angle is –480°.

    d) 11π 7π2π2 2

    11π π6π 2 2

    For an angle of 11π2

    , one positive coterminal angle is 7π2

    and one negative coterminal

    angle is π2

    .

    e) –205° + 360° = 155° –205° – 360° = –565° For an angle of –205°, one positive coterminal angle is 155° and one negative coterminal angle is –565°. f) 7.8 – 2π ≈ 1.5 7.8 – 4π ≈ –4.8 For an angle of –7.8, one positive coterminal angle is 1.5 and one negative coterminal angle is –4.8.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 5 of 85

    Section 4.1 Page 176 Question 8

    a) The angles 5π6

    and 17π6

    are coterminal because

    5π 5π 12π2π6 6 6

    17π6

    b) The angles 5π2

    and 17π6

    are not coterminal because 5π2

    is coterminal with π2

    which

    falls on the positive y-axis, while 17π6

    is coterminal with 5π6

    , which is in quadrant II.

    c) The angles 410° and –410° are not coterminal because 410° is coterminal with 50° and so is in quadrant I, while –410° is coterminal with 310° and is in quadrant IV. d) The angles 227° and –493° are coterminal because –493° is coterminal with –493° + 2(360°) which is 227°. Section 4.1 Page 176 Question 9 a) The coterminal angles for 135° are 135° ± (360°)n, where n is any natural number.

    b) The coterminal angles for π2

    are π 2π2

    n , where n is any natural number.

    c) The coterminal angles for –200° are –200° ± (360°)n, where n is any natural number. d) The coterminal angles for 10 radians are 10 ± 2πn, where n is any natural number. Section 4.1 Page 176 Question 10 Example: Choose –45°. –45° = –45° + 360° = 315° In general, all angles coterminal with –45° are given by –45° ± (360°)n, where n is any natural number.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 6 of 85

    Section 4.1 Page 176 Question 11 a) 65° + 360° = 425° In the domain 0° ≤ θ < 720°, the angle 425° is coterminal with 65°. b) –40° + 360° = 320° In the domain –180° ≤ θ < 360°, the angle 320° is coterminal with –40°. c) –40° + 360° = 320° –40° – 360° = –400° –40° + 2(360°) = 680° In the domain –720° ≤ θ < 720°, the angles –400°, 320°, and 680° are coterminal with –40°.

    d) 3π 5π2π4 4

    In the domain –2π ≤ θ < 2π, the angle 5π4

    is coterminal with 3π4

    .

    e) 11π 23π2π6 6

    11π π2π6 6

    11π 13π4π6 6

    In the domain –4π ≤ θ < 4π, the angles 23π6

    , π6

    and 13π

    6 are coterminal with 11π

    6 .

    f) 7π π2π3 3 7π 5π4π

    3 3

    In the domain, –2π ≤ θ < 4π, the angles π3

    and 5π3

    are coterminal with 7π3

    .

    g) 2.4 – 2π ≈ –3.9 In the domain –2π ≤ θ < 2π, the angle –3.9 is coterminal with 2.4. h) –7.2 + 2π ≈ –0.9 –7.2 + 4π ≈ 5.4 –7.2 – 2π ≈ –13.5 (outside specified domain) In the domain –4π ≤ θ < 2π, the angles –0.9 and 5.4 are coterminal with –7.2.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 7 of 85

    Section 4.1 Page 176 Question 12 a) Use a proportion with r = 9.5 and central angle 1.4 radians.

    arc length central angle=circumference complete rotation

    arc length2π( ) 2π

    arc length 1.4(9.

    1.49.5

    5)13.3

    The arc length is 13.3 cm. b) Use the formula a = θr, with r = 1.37 and θ = 3.5. a = 3.5(1.37) = 4.795 The arc length is 4.80 m, to the nearest hundredth of a metre. c) Use a proportion with r = 7 and central angle 130°.

    arc length central angle=circumference complete rotation

    arc length2π( ) 360

    13(14π)ar

    13

    c le

    07

    ngth 36

    15.88

    The arc length is 15.88 cm, to the nearest hundredth of a centimetre. d) Use a proportion with r = 6.25 and central angle 282°.

    arc length central angle=circumference complete rotation

    arc length2π( ) 360

    282(12.5π)arc l

    2826.2

    ength 3

    5

    6030.76

    The arc length is 30.76 in., to the nearest hundredth of an inch.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 8 of 85

    Section 4.1 Page 176 Question 13 a) Use the formula a = θr, with a = 9 and r = 4. 9 = θ(4)

    94

    = θ

    2.25 = θ The central angle is 2.25 radians. b) Use the formula a = θr, with θ = 1.22 and r = 9. a = 1.22(9) = 10.98 The arc length is 10.98 ft. c) Use the formula a = θr, with a = 15 and θ = 3.93. 15 = 3.93r

    153.93

    = r

    3.82 ≈ r The radius is 3.82 cm, to the nearest hundredth of a centimetre. d) Use a proportion with r = 7 and central angle 140°.

    arc length central angle=circumference complete rotation

    arc length2π( ) 360

    14(14π)ar

    14

    c le

    07

    ngth 36

    17.10

    The arc length is 17.10 m, to the nearest hundredth of a metre. Section 4.1 Page 176 Question 14

    a) Use the formula a = θr, with r = 5 and θ = 5π3

    .

    5π 5

    25π3

    26. 8

    3

    1

    a

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 9 of 85

    The arc length of the sector watered is 25π3

    m or 26.18 m, to the nearest tenth of a metre.

    b) Use a proportion with r = 5 and central angle 5π3

    .

    2

    area of sector central angle=area of circle complete rot

    5π3

    ation

    area of sectorπ( ) 2π

    5(25π)area of sector 6

    5

    The area of the sector watered is 125π6

    , or approximately 65.45 m2.

    c) The sprinkler makes one revolution every 15 s, so in 2 min it will make 8 revolutions. In 2 min the sprinkler will rotate through 8(2π) radians, which is 16π radians, or 8(360°) which is 2880°. Section 4.1 Page 177 Question 15 a) One revolution in 24 h is the same as: 36024 per hour, which is 15°/h; or π radians in 12 h, which is π

    12 radians/h

    b) 1000 rpm = 1000(2π) radians/min

    = 1000(2π)60

    radians/s

    The angular velocity of the motor is 100π3

    radians per second.

    c) 10 revolutions in 4 s is 10(360 )4

    or 900° per second.

    In one minute, this is 60(900°) or 54 000°. The angular velocity of the bicycle wheel is 54 000°/min.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 10 of 85

    Section 4.1 Page 177 Question 16 a) Use the formula a = θr, with a = 170 and r = 72. 170 = θ(72) 17072

    = θ

    2.36 ≈ θ The central angle of the cable swing is 2.36 radians, to the nearest hundredth of a radian.

    b) 2.36 radians 1802.36π

    135.3

    The measure of the central angle is 135.3°, to the nearest tenth of a degree. Section 4.1 Page 177 Question 17

    Revolutions Degrees Radians

    a) 1 rev 360° 2π

    b) 0.75 rev 270° 3π2

    or 4.7

    c) 0.4 rev 150° 5π6

    d) –0.3 rev –97.4° –1.7

    e) –0.1 rev –40° 2π9

    or –0.7

    f) 0.7 rev 252° 7π5

    or 4.4

    g) –3.25 rev –1170° 13π2

    or –20.4

    h) 2318

    or 1.3 rev 460° 23π9

    or 8.0

    i) 316

    or –0.2 rev –67.5° 3π8

    Section 4.1 Page 177 Question 18 Joran’s answer includes the given angle, obtained when n = 0. Jasmine’s answer is better as it excludes the actual given angle and just generates all positive and negative coterminal angles.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 11 of 85

    Section 4.1 Page 177 Question 19 a) 360° = 400 grads

    So, 1° = 400360

    grads

    Then, 50° 40050

    10

    3609

    5009

    55.6 grads

    b) Use the equivalence 360° = 400 grads. To convert from degrees to gradians, multiply

    the number of degrees by 400360

    . To convert from gradians to degrees, multiply the

    number of gradians by 360400

    .

    c) The gradian was developed in France along with the metric system. A right angle, or 90° is 100 gradians and so fractions of a right angle can be expressed in decimal form in gradians. Section 4.1 Page 178 Question 20 a) The central angle is 62.45° – 49.63°, or 12.82°.

    b) Use a proportion with r = 6400 and central angle 12.82°.

    arc length central angle=circumference complete rotation

    arc length2π( ) 360

    12.82(1280π)arc le

    12.82640

    ngth 36

    143 . 1

    0

    2 0

    The distance between Yellowknife and Crowsnest Pass is 1432.01 km, to the nearest hundredth of a kilometre.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 12 of 85

    c) Example: Bowden and Aidrie are both approximately 114° W. Bowden is at 51.93° N and Airdrie is at 51.29° N. So, the central angle is 51.93° – 51.29° or 0.64°. Use a proportion with r = 6400 and central angle 0.64°.

    arc length central angle=circumference complete rotation

    arc length2π( ) 360

    0.64(1280π)arc

    0

    length 36

    71.

    .66400

    4

    4

    9

    The distance between Bowden and Aidrie is 71.49 km, to the nearest hundredth of a kilometre. Section 4.1 Page 178 Question 21

    a) 133.284 km/h 133.284(1000) m/min60

    2221.4 m/min

    Sam Whittingham’s speed, in the 200-m flying start race, was 2221.4 m/min. b) The bicycle wheel circumference is π(0.6) m.

    So, the number of wheel turns in 2221.4 m is 2221.40.6π

    .

    Then, the angular speed of the wheel is 2221.4 (2π)0.6π

    or approximately 7404.7 radians

    per minute. Section 4.1 Page 178 Question 22 Speed of the water wheel is 15 rpm or 15(3π) m/min. Convert the speed to kilometres per hour.

    15(3π) m/min 15(3π)(60) km/h1000

    8.5 km/h

    The speed of the water wheel is approximately 8.5 km/h.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 13 of 85

    Section 4.1 Page 178 Question 23 In one revolution about the sun, Earth travels 93 000 000(2π) miles. Convert 93 000 000(2π) miles in 365 days to a speed in miles per hour.

    This is 93 000 000(2π)365(24)

    miles per hour.

    The speed of Earth is approximately 66 705.05 mph. Section 4.1 Page 178 Question 24

    a) 69.375° 37569 601000

    69 22.569 22 30

    b) i) 40.875° 87540 601000

    40 52.540 52 30

    ii) 100.126° 126100 601000

    100 7.5656100 7 60

    100

    100 7 33.6

    iii) 14.565° 56514 601000

    14 33.9914 33 60

    10

    14 33 54

    iv) 80.385 38580 601000

    80 23.1180 23 60

    1080 23 6

    Section 4.1 Page 179 Question 25

    a) 3069 22 30 69 22 +60

    69 22.5

    22.56960

    69.375

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 14 of 85

    b) i) 3045 30 30 45 30 +60

    45 30.5

    30.54560

    45.508

    ii) 4572 15 45 72 15 +60

    72 15.75

    15.757260

    72.263

    iii) 15105 40 15 105 40 +60

    105 40.25

    40.2510560

    105.671

    iv) 28 10 28 10

    102860

    28.167

    Section 4.1 Page 179 Question 26 First, write an expression for the area of the sector. Use a proportion with central angle θ.

    2

    2

    area of sector θarea of circle 2π

    θ(π )area of sector2π

    θ2

    r

    r

    Next, derive an expression for the area of the triangle. The perpendicular height from the centre of the circle to AB will bisect the triangle into

    two congruent right triangles. The height will be rcos θ2

    . The length AB will be

    2r sin θ2

    .

    Area of triangle

    2

    2

    θ θ(2 sin )( cos )2 2

    2θ θsin cos2 2

    1 sin θ2

    r r

    r

    r

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 15 of 85

    Then, area of shaded segment = area of sector – area of triangle

    2

    2

    2

    θ 1 sinθ2 2

    (θ sinθ)2

    r r

    r

    Section 4.1 Page 179 Question 27 a) At 4:00, the minute hand is at 12 and the hour hand is at 4.

    Angle 4 (360 )12120

    The angle between the hand of a clock at 4:00 is 120°.

    b) At 4:10 the hour hand will have moved 10 3060

    , or 5° past the 4. The minute hand

    will have rotated 10 36060

    , or 60° from the 12. The angle between the hands at 4:10 is

    120° + 5° – 60°, or 65°. c) Example: The hands are at right angles to each other at 3:00 and at 9:00. d) The hand are at right angles twice between 4:00 and 5:00. Represent the time as 4 + x, where x is the number of minutes past the hour.

    Then, the hour hand will have moved 3060x

    . The minute hand will have rotated

    36060x

    . The angle between the hands is 90° when

    120° + 3060x

    – 360

    60x

    = 90°

    Solve the equation:

    120 + 2x – 6x = 90

    60 = 11x

    60115.5

    x

    x

    To the nearest minute, the hands will be at right angles to each other at 4:05. There is another such time, when the minute hand is beyond the hour hand. In this case, the minute hand is ahead of the hour hand, so the equation to solve is

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 16 of 85

    6x – 120 – 2x = 90

    112x = 210

    11x = 420 x ≈ 38 To the nearest minute, the hands will be at right angles to each other at 4:38. e) As shown in part d) above, one time occurs shortly after 4:05. Section 4.1 Page 179 Question C1 One revolution is 2π radians, which is approximately 6.28 radians. So, 6 radians is a little less than 360°. Section 4.1 Page 179 Question C2

    One degree is 1360

    th of a complete revolution; so it is a

    very small angle. On the other hand, 1 radian is an amount of rotation that cuts off an arc with length equal

    to the radius. It is a little less than 16

    th of a complete

    revolution.

    Section 4.1 Page 179 Question C3 a) 860° – 720° = 140° The reference angle is 180° – 140°, or 40°. An expression for all coterminal angles is 140° ± 360°n, n N. b) (–7 + 4π) rad ≈ 5.566 rad The reference angle is 2π – 5.566 rad, or 0.72 rad, to the nearest hundredth. An expression for all coterminal angles is 0.72 ± 2πn, n N.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 17 of 85

    Section 4.1 Page 179 Question C4 a)

    b)

    Section 4.1 Page 179 Question C5 a) x = 3

    b) y = x – 3

    Section 4.2 The Unit Circle Section 4.2 Page 186 Question 1 a) In x2 + y2 = r2, substitute r = 4. x2 + y2 = 16 b) In x2 + y2 = r2, substitute r = 3. x2 + y2 = 9 c) In x2 + y2 = r2, substitute r = 12. x2 + y2 = 144 d) In x2 + y2 = r2, substitute r = 2.6. x2 + y2 = 2.62 x2 + y2 = 6.76

    P(x, y)

    x

    y

    O

    r

    –3

    45° 0

    y

    x (3, 0) 0

    y

    x

    (3, 0)

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 18 of 85

    Section 4.2 Page 186 Question 2 A point is on the unit circle if x2 + y2 = 1.

    a) For 3 1,4 4

    ,

    2 23 1 9 14 4 16 16

    10 5 or 16 81

    Therefore, the point 3 1,4 4

    is not on

    the unit circle.

    b) For 5 7,8 8

    ,

    2 25 7 5 498 8 64 64

    54 164

    Therefore, the point 5 7,8 8

    is not on

    the unit circle.

    c) For 5 12,13 13

    ,

    2 25 12 25 14413 13 169 169

    1691691

    Therefore, the point 5 12,13 13

    is on

    the unit circle.

    d) For 4 3,5 5

    ,

    2 24 3 16 95 5 25 25

    25251

    Therefore, the point 4 3,5 5

    is on

    the unit circle.

    e) For 3 1,2 2

    ,

    2 23 1 3 12 2 4 4

    441

    Therefore, the point 3 1,2 2

    is on

    the unit circle.

    f) For 7 3,4 4

    ,

    2 27 3 7 94 4 16 16

    16161

    Therefore, the point 7 3,4 4

    is on the

    unit circle.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 19 of 85

    Section 4.2 Page 187 Question 3

    a) 2

    2

    2

    2

    1

    1116

    14

    1516

    154

    y

    y

    y

    y

    In quadrant I, y = 154

    .

    b) 2

    2

    2

    1

    419

    23

    53

    x

    x

    x

    In quadrant II, x = 53

    .

    c) 2

    2

    2

    2

    1

    49164

    78

    1564

    158

    y

    y

    y

    y

    In quadrant III, y = 158

    .

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 20 of 85

    d) 2

    2

    2

    2

    1

    25149

    57

    2449

    247

    x

    x

    x

    x

    In quadrant IV, x = 247

    or 2 6

    7.

    e) 2

    2

    2

    2

    1

    119

    8

    13

    98

    3

    x

    x

    x

    x

    For x < 0, x = 8 2 2 or 3 3

    .

    f) 2

    2

    2

    2

    1

    1441169

    25169

    513

    1213

    y

    y

    y

    y

    The point is not in quadrant I. Since it has a positive x-value, the point must be in quadrant IV

    with y = 513

    .

    Section 4.2 Page 187 Question 4 a) A rotation of π radians takes the terminal arm of the angle on to the x-axis to the left of the origin. So, P(π) = (–1, 0).

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 21 of 85

    b) A rotation of π2

    radians takes the terminal arm of the angle onto the y-axis below

    the origin.

    So, P π2

    = (0, –1).

    c) A rotation of π3

    radians takes the

    terminal arm of the angle into the first quadrant as shown.

    So, P π 1 3,3 2 2

    .

    d) A rotation of π6

    radians takes the

    terminal arm of the angle into quadrant IV as shown.

    So, P π 3 1,6 2 2

    .

    e) A rotation of 3π4

    radians takes the

    terminal arm of the angle into the second quadrant as shown.

    So, P 3π 2 2,4 2 2

    .

    f) A rotation of 7π

    4 radians takes the terminal arm of the angle into the first quadrant

    and is coterminal with π4

    .

    So, P 7π 2 2,4 2 2

    .

    g) A rotation of 4π is two complete turns and is coterminal with 0 radians. So, P(4π) = (1, 0).

    (1, 0)

    ,3 1

    2 2

    (1, 0)

    2 2,

    2 2

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 22 of 85

    h) A rotation of 5π2

    radians is one complete turn plus one-half turn and takes the

    terminal arm of the angle onto the y-axis above the origin.

    So, P 5π2

    = (0, 1).

    i) A rotation of 5π6

    radians takes the

    terminal arm of the angle into quadrant II as shown.

    So, P π 3 1,6 2 2

    .

    j) A rotation of 4π3

    radians takes the

    terminal arm of the angle into quadrant II,

    with reference angle π3

    as shown.

    So, P 4π 1 3,3 2 2

    .

    Section 4.2 Page 187 Question 5

    a) (0, –1) is on the y-axis, below the origin, so θ = 3π2

    .

    b) (1, 0) is on the x-axis, to the right of the origin, so θ = 0.

    c) 2 2,2 2

    is in quadrant I, and since x and y are equal the measure of the central

    angle is π4

    .

    d) 1 1,2 2

    is in quadrant II, and since x and y have equal length, the measure of the

    reference angle is π4

    . In quadrant II, θ = 3π4

    .

    (1, 0)

    ,1

    2 2

    3

    ,3 1

    2 2

    (1, 0)

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 23 of 85

    e) 1 3,2 2

    is in quadrant I and θ = π3

    as shown.

    f) 1 3,2 2

    is in quadrant IV, and so θ = 2π – π

    3, or 5π

    3.

    g) 3 1,2 2

    is in quadrant II as shown

    and θ = 5π6

    .

    h) 3 1,2 2

    is similar to part g), except the terminal arm is in quadrant III.

    So, θ = 7π6

    .

    i) 2 2,2 2

    is in quadrant III, and since x and y are equal the measure of the central

    angle is π + π4

    . So, θ = 5π4

    .

    j) (–1, 0) in on the x-axis, to the left of the origin. So, θ = π. Section 4.2 Page 187 Question 6

    If P(θ) = 3 1,2 2

    , then θ is in

    quadrant II as shown. One positive

    measure for θ is 5π6

    and a coterminal

    negative angle is 5π 7π2π, or 6 6 .

    (1, 0)

    ,3 1

    2 2

    (1, 0)

    ,3 1

    2 2

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 24 of 85

    Section 4.2 Page 187 Question 7

    a) Example: Choose θ = π3

    , then

    θ + π = π 4ππ, or 3 3 .

    π 1 3P ,3 2 2

    and

    4π 1 3P ,3 2 2

    b) Example: Choose θ = 3π4

    , then

    θ + π = 3π 7ππ, or 4 4 .

    3π 1 1 2 2P , , or ,4 2 22 2

    and 7π 1 1 2 2P , , or ,4 2 22 2

    Section 4.2 Page 187 Question 8 Point

    Step 2:

    14

    turn

    Step 3:

    14

    turn

    Step 4:

    Description

    Diagram

    P(0) = (1, 0)

    πP

    20, 1

    πP

    20, 1

    x- and y-values change places and take appropriate signs for the new quadrant

    πP

    3

    1 3,

    2 2

    π π P3 25πP6

    3 1, 2 2

    π π P3 2πP6

    3 1, 2 2

    x- and y-values change places and take appropriate signs for the new quadrant

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 25 of 85

    5πP

    3

    1 3,

    2 2

    5π π P3 2πP6

    3 1, 2 2

    5π π P3 27πP6

    3 1, 2 2

    x- and y-values change places and take appropriate signs for the new quadrant

    Step 4: Concluding diagram.

    Section 4.2 Page 188 Question 9 a) The diagram shows a unit circle: x2 + y2 = 1. b) C is related to B by a 90° rotation, so as shown in the previous question, the coordinates switch and the signs are adjusted. In quadrant I, both coordinates are positive. So the coordinates of B

    are 5 2,3 3

    .

    c)

    πAC AB2

    πAC θ2

    d) If P(θ) = B, then πP θ2

    is related by a rotation of π2

    clockwise, which puts it in

    quadrant IV. e) The maximum value for the x-coordinates or the y-coordinates is 1. The minimum value for the x-coordinates or the y-coordinates is –1. Section 4.2 Page 188 Question 10 a) Mya is correct, because in quadrant I the x-coordinates start at 1 for an angle of 0° and decrease to a minimum of 0 for an angle of 90°.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 26 of 85

    b) Check Mya’s answer by substituting x = 0.807 and y = 0.348 751 into the equation for the unit circle, x2 + y2 = 1. If the sum of the squares on the left side is not equal to 1, then Mya has made an error and her calculation needs checking. Observe that Mya forgot to the take the square root. Left Side = (0.807)2 + (0.348 751)2 = 0.772 876 26 ≠ 1 Recalculate: when x = 0.807 (0.807)2 + y2 = 1 y2 = 1 – (0.807)2 y = 21 (0.807) y ≈ 0.590 551 c) Substitute y = 0.2571 in x2 + y2 = 1 x2 + (0.2571)2 = 1 x2 = 1 – (0.2571)2 x = 21 (0.2571) x ≈ 0.9664 Section 4.2 Page 188 Question 11 a)

    b) The denominators of the coordinates are all 2. c) The numerators of the x-coordinates are decreasing as P(θ) increases, while the y-coordinates are increasing. This makes sense, in the quadrant I, because the terminal

    arm is getting closer to the y-axis as the angle increases. At P π4

    the x-coordinate and

    the y-coordinate are the same.

    d) Square roots are derived from the special right triangles with acute angles π4

    , π4

    , and

    π3

    , π6

    .

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 27 of 85

    e) Example. Remember the patterns of side ratios for the special right triangles and that

    for angles less than π4

    , the x-coordinate is greater than the y-coordinate.

    Section 4.2 Page 188 Question 12 a) The interval –2π ≤ θ < 4π represents three rotations around the unit circle: one complete clockwise rotation starting at 0 and two complete counterclockwise rotations. For every point on the unit circle there will be three coterminal angles in this interval.

    b) If P(θ) = 1 3,2 2

    , then θ is in quadrant II.

    In the interval –2π ≤ θ < 0, θ = 4π3

    .

    In the interval 0 ≤ θ < 2π, θ = 2π3

    .

    In the interval 2π ≤ θ < 4π,

    θ = 2π 8π2π , or 3 3

    .

    c) The terminal arm of the three angles is in the same position on the unit circle and all three angles have the same reference angle. The angles are “coterminal”. Section 4.2 Page 188 Question 13

    a) If P(θ) = 1 2 2,3 3

    , then θ is an

    angle with terminal arm in quadrant III. The location of P, at the intersection of the terminal arm and the unit circle, is

    given by x = 13

    and y = –2 2

    3.

    b) θ terminates in quadrant III.

    c) πP θ2

    will be in quadrant IV and for a rotation of + π2

    the coordinates of P from

    part a) switch and the signs are adjusted for quadrant IV. π 2 2 1P θ ,2 3 3

    (1, 0)

    ,3 1

    2 2

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 28 of 85

    d) πP θ2

    will be in quadrant II and for a rotation of – π2

    the coordinates of P from

    part a) switch and the signs are adjusted for quadrant II. π 2 2 1P θ ,2 3 3

    Section 4.2 Page 189 Question 14 π units is a length. On the unit circle it is the arc of an angle from (1, 0) to (–1, 0). π square units is an area. It is the area of the unit circle because when r = 1 in A = πr2 the area is A = π(1)2, or π square units.

    Section 4.2 Page 189 Question 15 a) Since ABCD is a rectangle, opposite sides are the same length. So, the coordinates of the other vertices are: B(–a, b) C(–a, –b) D(a, –b)

    b) i) Since π is half a complete rotation, θ + π will have terminal arm OC. The angle will pass through C. ii) Similarly, – π is half a complete rotation, in the clockwise direction, so θ – π will have terminal arm OC. The angle will pass through C. iii) –θ is the FOD. A rotation of π from OD will have terminal arm OB. The angle –θ + π will pass through B. iv) A rotation of –π from OD will have terminal arm OB. The angle –θ – π will pass through B. c) The answers would be the same because, since the angles are expressed in radians, arc FA is the same as θ.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 29 of 85

    Section 4.2 Page 189 Question 16

    a) In a counterclockwise direction, arc SG is created by a rotation of 5π4

    .

    The arc length of SG is 5π4

    .

    b) P 13π2

    represents a point on the unit circle

    obtained by rotating an angle in standard position through 3 complete rotations plus half a rotation.

    We know this because 13π 12π π π, or 6π2 2 2 2

    and 6π is 3 complete rotations of 2π each. The point on the unit circle corresponding to this amount of rotation is A.

    c) P(5) is in quadrant IV. Since P 3π2

    is approximately P(4.71) and P(2π) is

    approximately P(6.28), so P(5) is between C and D. Section 4.2 Page 189 Question 17 a) y = –3x x2 + y2 = 1 Substitute from into . x2 + (–3x)2 = 1 10x2 = 1

    x = 110

    Substitute in to find the corresponding y-values 110

    3

    3 3 or 10 10

    y

    y y

    The points of intersection are 1 3, , or (0.1 10, 0.3 10)10 10

    , and

    1 3, , or ( 0.1 10,0.3 10)10 10

    .

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 30 of 85

    b) The first point of intersection above is in quadrant IV and the second point is a rotation of π away, in quadrant II. Since the first coordinate is cos θ, the measure of reference angle θ is approximately 1.25 radians.

    Section 4.2 Page 189 Question 18 a) First, use the Pythagorean theorem to determine the length of the hypotenuse OA. OA2 = 52 + 22 OA = 29 Next, compare sides of the similar right triangles.

    51 29

    529

    x

    x

    21 29

    529

    y

    y

    The exact coordinates of P(θ) are 5 2,29 29

    .

    b) The radius of the larger circle passing through A is the length of OA found in part a). It is 29 . c) The equation for the larger circle passing through A is x2 + y2 = 29. Section 4.2 Page 190 Question 19 Consider P(θ) = (x, y). In the unit circle, the hypotenuse is 1.

    Then, cos θ adjacenthypotenuse

    1x

    x

    and sin θ oppositehypotenuse

    1y

    y

    So, P(θ) = (cos θ, sin θ).

    P(θ)

    y

    A(5, 2)

    x

    2

    O x

    y

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 31 of 85

    Section 4.2 Page 190 Question 20

    a) 2 2,

    2 2

    are the coordinates of a point on the unit circle in quadrant I, describing a

    rotation of π4

    . The equivalent polar coordinates are π1, 4

    .

    b) 3 1,2 3

    are the coordinates of a point on a circle in quadrant III.

    First, determine the radius of the circle: 2 2

    2

    2

    2

    3 12 3

    3 14 9

    3136

    316

    r

    r

    r

    r

    Next, determine the angle measure.

    1 33 2

    tanθ

    1 23 32

    2 3

    yx

    So, θ ≈ 0.367 This is the measure of the reference angle, so in quadrant III the angle is π + 0.367 or 3.509.

    The equivalent polar coordinates are 31 , 3.5096

    .

    c) (2, 2) are the coordinates of a point on a circle in quadrant I. Since x = y this is a

    rotation of π4

    .

    Determine the radius: 22 + 22 = r2 8 = r2

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 32 of 85

    r = 2 2

    The equivalent polar coordinates are π2 2, 4

    .

    d) (4, –3) are the coordinates of a point on a circle in quadrant IV. First, determine the radius: 42 + (–3)2 = r2 25 = r2 r = 5 Next determine the angle measure.

    3tanθ4

    θ 0.644

    This is the measure of the reference angle, so in quadrant IV the angle is 2π – 0.644 or 5.640. The equivalent polar coordinates are (5, 5.640). Section 4.2 Page 190 Question C1 a)

    b)

    c) Using the special right triangle for π6

    , the vertex in quadrant I has coordinates

    3 1,2 2

    . Then adjust the coordinates for the vertices of the rectangle in other quadrants.

    In quadrant II, P 5π6

    = 3 1,2 2

    .

    In quadrant III, P 7π6

    = 3 1,2 2

    .

    In quadrant IV, P 11π6

    = 3 1,2 2

    .

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 33 of 85

    d) Example: Divide the unit circle into eighths to mark off multiples of π4

    . Use the ratio

    of sides of the special isosceles right triangle with hypotenuse 1 ( 22

    : 22

    : 1) adjusting

    the signs of coordinates for each quadrant. Think of hours on a clock face to mark off

    multiples of π6

    . Then, use the ratio of sides of the special right triangle with hypotenuse 1

    ( 32

    : 12

    : 1) adjusting the signs of coordinates for each quadrant.

    Section 4.2 Page 190 Question C2 a) Let n represent the measure of BOA. Then, ABO = BAO = 2n. Arc AB is the same as the measure of BAO, in radians. Use the angle sum of a triangle. n + 2n + 2n = π 5n = π

    n = π5

    The measure of arc AB is π5

    .

    b) If P(C) = P πB2

    , then C must be in quadrant II.

    The measure of arc AC, and of COA, is π π 7π5 2 10

    Then in ∆OAC,

    CAO + ACO = π – 7π10

    = 3π10

    Since CAO = ACO, CAO = 3π20

    Section 4.2 Page 190 Question C3 a) x2 + y2 = r2

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 34 of 85

    b) Example: If the centre of the circle is moved to (h, k), then a new right triangle can be used to determine the equation. Its horizontal length will be (x – h), and its vertical side will be (y – k). Then, using the Pythagorean theorem (x – h)2 + (y – k)2 = r2.

    Section 4.2 Page 190 Question C4 a) Area of circle = π(12) Area of the square = 22

    Percent of paper cut off = 4 π 1004

    ≈ 21.5% b) Circumference : perimeter of square = 2π : 8 = π : 4 4.3 Trigonometric Ratios Section 4.3 Page 201 Question 1

    a) sin 45°

    22

    yr

    b) tan 30° =

    1 3 or 33

    123

    2

    xy

    1

    2

    3

    2

    30°

    1

    (h, k)

    (x, y)

    2

    2

    45°

    1

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 35 of 85

    c) 3πcos4

    12

    22

    2

    xr

    d) 7πcot6

    123

    23

    yx

    e) Refer to the diagram in part d), because 7π6

    = 210°.

    csc 210°

    112

    2

    ry

    f) sec (–240°)

    112

    2

    rx

    g) A point on the terminal arm of 3π2

    is (0, –1).

    Therefore,

    10

    3πtan2

    undefined

    yx

    –240°

    1

    1

    2

    3

    2

    6

    1

    2

    3

    1

    2

    1

    2

    2

    2

    2 3π

    4

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 36 of 85

    h) A point on the terminal arm of π is (–1, 0).

    Therefore, secπ

    1

    11

    rx

    i) cot (–120°)

    123

    21 3 or

    33

    xy

    j) A rotation of 390° is coterminal with 30°. So, cos 390° = cos 30°

    = 32

    k) 5πsin3

    13

    32

    2

    yr

    l) A rotation of 495° is coterminal with 495° – 360°, or 135°. Therefore, csc 495° = csc 135°

    12

    22 or 22

    ry

    1

    2

    2

    2

    2

    1

    1

    2

    3

    25π

    3

    –120°

    1

    1

    2

    3

    2

    495º

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 37 of 85

    Section 4.3 Page 201 Question 2 Use a calculator. Verify that the sign is correct for the quadrant.

    a) cos 47° ≈ 0.68 b) cot 160° 1tan1602.75

    c) sec 15° 1cos151.04

    d) csc 4.71 1sin 4.711.00

    e) sin 5 ≈ –0.96 f) tan 0.94 ≈ 1.37

    g) 5πsin 0.787 h) tan 6.9 ≈ 0.71

    i) cos 302° ≈ 0.53 j) 11πsin 0.9719

    k) cot 6 1tan 63.44

    l) sec (–270°) 1cos( 270 )10undefined

    Section 4.3 Page 202 Question 3 The diagram shows a memory aid for the ratios that are positive, i.e. greater than 0, in each quadrant. a) cos θ > 0 in quadrants I and IV b) tan θ < 0 in quadrants II and IV c) sin θ < 0 in quadrants III and IV d) sin θ > 0 in quadrants I and II, cot θ < 0 in quadrants II and IV, so both conditions are true in quadrant II. e) cos θ < 0 in quadrants II and III, csc θ > 0 in quadrants I and II, so both conditions are true in quadrant II.

    All

    CosineCos

    Sin

    Tan

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 38 of 85

    f) sec θ > 0 in quadrants I and IV, tan θ > 0 in quadrants I and III, so both conditions are true in quadrant I. Section 4.3 Page 202 Question 4 a) 250° = 180° + 70°, so 250° is in quadrant III. In quadrant III, sine is negative. So, sin 250° = –sin 70°. b) 290° = 360° – 70°, so 290° is in quadrant IV. In quadrant IV, tangent is negative. So, tan 290° = –tan 70°. c) 135° = 180° – 45°, so 135° is in quadrant II. In quadrant II, cosine and secant are negative. So, sec 135° = –sec 45°. d) 4 radians is in quadrant III and its reference angle is 4 – π. In quadrant III, cosine is negative. So, cos 4 = –cos (4 – π). e) 3 radians is in quadrant II and its reference angle is π – 3. In quadrant II, sine and cosecant are positive. So, csc 3 = csc (π – 3). f) 4.95 radians is in quadrant III and its reference angle is 4.95 – π. In quadrant III, tangent and cotangent are positive. So, cot 4.95 = cot (4.95 – π). Section 4.3 Page 202 Question 5 a) (3, 5) is in quadrant I.

    Use

    5

    anθ

    3

    t yx

    Then, the reference angle is θ ≈ 1.03. A negative coterminal angle is 1.03 – 2π ≈ –5.25.

    b) (–2, –1) is in quadrant III.

    Use

    1

    tanθ

    or 0.2

    5

    yx

    Then, the reference angle is θ ≈ 0.4636…. One positive angle, in quadrant III, is π + θ or approximately 3.61. A negative coterminal angle is 3.61 – 2π ≈ –2.68.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 39 of 85

    c) (–3, 2) is in quadrant II.

    Use

    2

    anθ

    3

    t yx

    Then, a reference angle is θ ≈ 0.588…. One positive angle, in quadrant II, is π – θ or approximately 2.55. A negative coterminal angle is –π – θ or approximately –3.73.

    d) (5, –2) is in quadrant IV.

    Use

    2

    anθ

    5

    t yx

    Then, a reference angle is θ ≈ –0.38. This angle is in quadrant IV. A positive coterminal angle is 2π + θ or approximately 5.90.

    Section 4.3 Page 202 Question 6

    a) cosθ xr

    . 300° is in quadrant IV, so the x-coordinate of a point on the terminal arm is

    positive. Therefore, cos 300° is positive.

    b) sin θ yr

    . 4 radians is in quadrant III, so the y-coordinate of a point on the terminal

    arm is negative. Therefore, sin 4 is negative.

    c) cotθ xy

    . 156° is in quadrant II, so the x-coordinate of a point on the terminal arm is

    negative and the y-coordinate is positive. Therefore, cot 156° is negative.

    d) cscθ ry

    . –235° is in quadrant II, so the y-coordinate of a point on the terminal arm is

    positive. Therefore, cos 300° is positive.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 40 of 85

    e) tan θ yx

    . 13π6

    is coterminal with π6

    and is in quadrant I, so the x-coordinate of a

    point on the terminal arm is positive and the y-coordinate is positive. Therefore, 13πtan6

    is positive.

    f) secθ rx

    . 17π3

    is coterminal with 5π3

    and is in quadrant IV, so the x-coordinate of a

    point on the terminal arm is positive. Therefore, 17πsec3

    is positive.

    Section 4.3 Page 202 Question 7 a) sin–1 0.2 ≈ 0.2014 This means that an angle of 0.2014 radians has a sine ratio of 0.2. b) tan–1 7 ≈ 1.4289 This means that an angle of 1.4289 radians has a tangent ratio of 7.

    c) sec 450° = 1cos 450

    which is undefined because 450° is coterminal with 90° and

    cos 90° = 0.

    d) cot (–180°) = 1tan( 180 )

    which is undefined because (–180°) is coterminal with 180°

    and tan 180° = 0. Section 4.3 Page 202 Question 8

    a) Since P(θ) = 3 ,5

    y

    lies on the unit circle,

    x2 + y2 = 1 2

    2

    2

    2

    1

    9125

    1625

    4

    35

    5

    y

    y

    y

    y

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 41 of 85

    For P(θ) to be in quadrant IV, y must be negative. So, y = 45

    .

    b) tan θ

    45

    3543

    yx

    c) csc θ

    145541.25

    ry

    Section 4.3 Page 202 Question 9

    a) cos 60° + sin 30° 1 12 21

    b) (sec 45°)2 2

    2

    1cos 45

    112

    112

    2

    c) 5π 5π cos sec3 3

    5π 1cos5π3 cos3

    1

    d) (tan 60°)2 – (sec 60°)2

    2

    2 13 12

    3 41

    e) 2 2

    2 2

    7π 7π cos sin4 4

    1 12 2

    1 12 21

    f) 2 2

    2

    5π 5πcot 1 tan6 6

    113

    113

    3

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 42 of 85

    Section 4.3 Page 202 Question 10

    a) sin θ 12

    , 0 ≤ θ < 2π, means that θ is in

    quadrant III or IV. The reference angle is π6

    .

    In quadrant III, θ = π + π6

    , or 7π6

    .

    In quadrant IV, θ = 2π – π6

    , or 11π6

    .

    b) cot θ = 1, –π ≤ θ < 2π, means that θ is in quadrant I

    or III. The reference angle is π4

    .

    In quadrant I, θ = π4

    .

    In quadrant III, for a positive rotation, θ = π + π4

    , or

    5π4

    . There is also a negative angle in the given domain

    that falls in quadrant III; θ = –π + π4

    , or 3π4

    .

    c) sec θ = 2, –180° ≤ θ < 90°; cos θ = 12

    means that θ is in quadrant I or IV. The reference angle is 60°. In quadrant I, θ = 60°. The negative rotation that is in quadrant IV is –60°. d) If cos2 θ = 1, then cos θ = ±1. In the domain –360° ≤ θ < 360°, θ = 0°, 180°, –180°, –360°.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 43 of 85

    Section 4.3 Page 202 Question 11 a) cos θ = 0.42, gives a reference angle of θ ≈ 1.14. Cosine is positive in quadrants I and IV. So, in the domain –π ≤ θ ≤ π, θ ≈ 1.14 in quadrant I and θ ≈ –1.14 in quadrant IV.

    b) tan θ = –4.87, gives a value of θ ≈ –1.37. Tangent is negative in quadrants II and

    IV. So, in the domain π θ π2

    ,

    In quadrant II, θ ≈ π – 1.37, so θ ≈ 1.77. In quadrant IV, θ ≈ –1.37.

    c) csc θ = 4.87, means sin θ = 1 ,or sin θ 0.20534.87

    .

    This gives a value of θ ≈ 11.85°. Sine is positive in quadrants I and II. Consider the domain –360° ≤ θ < 180°. For positive rotations: in quadrant I, θ ≈ 11.85° and in quadrant II, θ ≈ 180° – 11.85°, or 168.15°. For negative rotations: in quadrant I, θ ≈ 11.85° – 360°, or –348.15°, and in quadrant II, –180° – 11.85°, or –191.85°.

    d) cot θ = 1.5, means tan θ = 1 ,or tan θ 0.66661.5

    .

    This gives a value of θ ≈ 33.69°. Tangent, and cotangent, is positive in quadrants I and III. Consider the domain –180° ≤ θ < 360°. For positive rotations: in quadrant I, θ ≈ 33.69° and in quadrant III, θ ≈ 33.69° + 180° or 213.69°. For negative rotations: in quadrant III, θ ≈ –180° + 33.69° or –146.31°.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 44 of 85

    Section 4.3 Page 202 Question 12

    a) Given 3sin θ5

    , and π θ π2 , then θ must

    be in quadrant II, as shown. The x-coordinate must be 3, since this is a 3-4-5 right triangle. The other five trigonometric ratios are:

    cosθ

    45

    xr

    tanθ

    34

    yx

    cscθ

    53

    ry

    secθ

    54

    rx

    cotθ

    43

    xy

    b) Given 2 2cosθ3

    , and 3ππ θ

    2 , then θ must be in quadrant II or III, with

    x = 2 2 and r = 3, as shown. Determine the y-coordinate. x2 + y2 = r2

    2 2 2

    22 2

    2

    2 2 3

    8 91

    x y r

    y

    yy

    The other five trigonometric ratios are:

    sinθ

    13

    yr

    tanθ

    12 2

    24

    yx

    cscθ 3 3secθ2 23 2

    4

    cotθ 2 2

    3

    2 2

    3

    3

    –4

    5

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 45 of 85

    c) Given 2tanθ3

    and –360° < θ < 180°, θ is in quadrant I or III, as shown.

    Determine the measure of r. 2 2 2

    2 2 2

    213

    13

    3 2x y r

    rr

    r

    sinθ

    213

    yr

    cosθ

    313

    xr

    13cscθ2

    13secθ3

    3cot θ2

    d) Given 4 3 3 3secθ , cosθ or 3 44 3

    , and –180° ≤ θ ≤ 180°, then θ is in quadrant

    I or IV, with x = 3 and r = 4 3 . Determine the y-coordinate.

    2 2 2

    22 2

    2

    3 4 3

    9 48

    39

    x y r

    y

    y

    y

    sinθ

    39 13 or 44 3

    yr

    tanθ

    393

    yx

    4 3 4 13cscθ or 1339

    3 39cotθ or 1339

    Section 4.3 Page 203 Question 13 Sketch the angle with point B(–2, –3) on its terminal arm. It is in quadrant III. Use the Pythagorean theorem to calculate the measure of r in the right triangle with x = –2 and y = – 3. The measure of r is 13 . The exact value of cos θ can be determined as

    4 3

    3

    r

    3

    2

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 46 of 85

    cosθ

    2 2 13 or1313

    xr

    Section 4.3 Page 203 Question 14

    a) 4900 22013360 360

    , so the angle of 4900° is 13 complete revolutions plus 220°.

    b) 220° puts the terminal arm in quadrant III. c) Since 220° is 40° past 180°, the reference angle is 40°. d) sin 4900° ≈ –0.643, cos 4900° ≈ –0.766, tan 4900° ≈ 0.839, csc 4900° ≈ –1.556, sec 4900° ≈ –1.305, cot 4900° ≈ 1.192 Section 4.3 Page 203 Question 15

    a) sin (cos–1 0.6) = 45

    or 0.8.

    To evaluate cos–1 0.6 find an angle whose cosine is 0.6. In other words, cos θ = xr

    = 610

    or 35

    . This means θ can be determined using a 3-4-5 right triangle. Then, the sine of this

    angle is sin θ = yr

    = 45

    .

    b) cos (sin–1 0.6) is very similar to the result in part a), the only difference being the orientation of the 3-4-5 right triangle. So, the positive value of cos (sin–1 0.6) is 0.8. Section 4.3 Page 203 Question 16 a) Jason is not correct. He used degree mode, when he should have chosen radian mode.

    b) First choose radian mode. Determine cos 40π7

    and then use the reciprocal key to

    determine sec 40π7

    . The correct answer is approximately 1.603 875 472.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 47 of 85

    Section 4.3 Page 203 Question 17 a) sin 1 ≈ 0.841, sin 2 ≈ 0.909, sin 3 ≈ 0.141, sin 4 ≈ –0.757 So the values in increasing order are sin 4, sin 3, sin 1, sin 2. b) 4 radians is in quadrant III, so sin 4 is negative. 3 radians is very close to π, so close to 0, but positive. The value of π/2 is approximately 1.57, so sin 2 is closer to the y-axis, where sine has value 1, than is sin 1. c) Cosine uses the x-coordinates which increase from left to right on the diagram. So the order is cos 3, cos 4, cos 2, cos 1.

    Check: cos 3 ≈ –0.999, cos 4 ≈ –0.654, cos 2 ≈ –0.416, cos 1 ≈ 0.540 Section 4.3 Page 203 Question 18 a) As P moves around the wheel it can move from closest to the piston at (1, 0) to furthest away at (–1, 0). So the maximum distance that Q can move is 2 units. b) At a speed of rotation is 1 radian/s, after 1 min the wheel will have travelled through 60 radians. 60 9.552π

    The wheel will have made 9 complete turns plus 0.55 of the next turn. This will put P in quadrant III at an angle of rotation of (0.55)2π or approximately 3.451 33 radians.

    c) After 1 s, P will have moved 1 radian and pulled Q to the left a distance of 1 – cos 1, or 0.46 units, to the nearest hundredth.

    O 1 – x x

    1

    1

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 48 of 85

    Section 4.3 Page 203 Question 19 a) A(–3, 4), domain 0 < θ ≤ 4π

    Using the coordinates of A, 4tan θ3

    .

    The reference angle is θ ≈ 0.93. A(–3, 4) is in quadrant II, so in the domain 0 < θ ≤ 4π, θ ≈ π – 0.93, or 2.21, and θ ≈ 2π + 2.21, or 8.50. b) B(5, –1), –360° ≤ θ < 360°

    Using the coordinates of B, 1tanθ 0.25

    .

    The reference angle is θ ≈ 11.31°. B(5, –1) is in quadrant IV, so in the domain –360° ≤ θ < 360°: for negative rotations: θ ≈ –11.31° for positive rotations: θ ≈ 360° – 11.31°, or 348.69°.

    c) C(–2, –3), domain 3π 7πθ2 2

    Using the coordinates of C, 3tan θ 1.52

    .

    The reference angle is θ ≈ 0.98.

    C(–2, –3) is in quadrant III, so in the domain 3π 7πθ2 2

    :

    for negative rotations: θ ≈ –π + 0.98, or –2.16 for positive rotations: θ ≈ π + 0.98, or 4.12 and 3π + 0.98, or 10.41 Section 4.3 Page 203 Question 20

    ∆BCD is a 30°-60°-90° triangle, so it sides have the proportions shown. ∆ABD is isosceles, so AD = BD = 2 units. Then in ∆ABC,

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 49 of 85

    BCtan15AC

    13 2

    Section 4.3 Page 204 Question 21 Note: The text answers show a rationale using degrees. This solution provides an alternate, using radians.

    The distance between (0, 5) and (5, 0) is π2

    .

    For the angle of rotation, θ, with x = 2.5,

    cos θ = 2.5 15 2

    . Therefore, θ = π3

    .

    So, moving from (0, 5) to the point on the curve where x = 2.5, the angle of rotation is

    – π6

    . This angle has an arc length of π6

    which is one-third of π2

    .

    Section 4.3 Page 204 Question 22 a)

    b) The new angle of rotation, πR6

    , has the same terminal arm as the angle in standard

    position P π3

    . So, π 1 3R ,6 2 2

    . The new angle of rotation, 5πR

    6

    , has the same

    terminal arm as the angle in standard position P 5π3

    . So, 5π 1 3R ,6 2 2

    .

    θ

    0 5

    5

    2.5

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 50 of 85

    c) The angles in the new system are related to angles in standard position by determining which quadrant the terminal arm is in and then determining the positive rotation to reach that terminal arm.

    π 1 3R ,6 2 2

    , 5π 1 3R ,

    6 2 2

    , 7π 1 3R ,6 2 2

    , 11π 1 3R ,

    6 2 2

    d) Bearings are measured clockwise from 0°. The new system is the same as bearings, except that bearings are measured in degrees. Section 4.3 Page 204 Question 23

    a) In ∆OBQ, cos θ = OB 1OQ OQ

    .

    Therefore, sec θ = 1 OQcosθ

    .

    b)

    In ∆OCD, ODC = θ (alternate angles).

    Then, sin θ = OC 1OD OD

    .

    So, csc θ = 1sin θ

    = OD.

    Similarly, cot θ = = CD.

    Section 4.3 Page 204 Question C1 a) Paula is correct, sine ratios are increasing in quadrant I.

    Examples: sin 0 = 0, sin π6

    = 0.5, sin π 2 0.7074 2 , sin π 3 0.866

    3 2 , sin π

    2 = 1.

    b) In quadrant II, sine is decreasing. In the unit circle, sine is given by the y-coordinate. As the end of the terminal arm moves past the y-axis, the y-coordinate decreases, from 1 to 0.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 51 of 85

    c) The sine ratio increases in quadrant IV. The y-coordinate has a minimum value of –1

    at 3π2

    and then its value increases to 0 at 0.

    Section 4.3 Page 204 Question C2 In this regular hexagon, the diagonals will intersect at the origin and each of the six angles at the center will be 60°. The vertex in quadrant I is at 60° and has coordinates

    1 3,2 2

    . The vertex in quadrant II is at 120° and has

    coordinates 1 3,2 2

    . The next vertex is at 180° and

    has coordinates (–1, 0). The vertex in quadrant III is at

    240° and has coordinates 1 3,2 2

    . The vertex in

    quadrant IV is at 300° and has coordinates 1 3,2 2

    .

    Section 4.3 Page 205 Question C3 a) If the coordinates of P are (x, y) then

    slope of OP

    tanθ

    yx

    b) Yes, the formula is true in all four quadrants. In quadrant II and IV the slope will be negative, as expected. c) An equation for the line OP, where O is the origin, is y = (tan θ)x.

    d) For any line, the equation is y = mx + b, where m is the slope and b is the y-intercept. The slope can be defined in terms of a unit circle as tan θ, if the circle is not centred at the origin then a vertical translation of b units is needed. The equation is y = (tan θ)x + b.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 52 of 85

    Section 4.3 Page 205 Question C4

    a) 1 4sin sin sinθ5

    45

    b) 1 4cos tan cosθ3

    35

    c) 1 3csc cos cscθ5

    1sinθ54

    Sine and cosecant are positive in quadrant II.

    d) 1 4sin tan sin(360 θ)3

    45

    Sine is negative in quadrant IV.

    4.4 Introduction to Trigonometric Equations Section 4.4 Page 211 Question 1 a) The given sine ratio is positive, so in the domain 0 ≤ θ < 2π, there will be two solutions, one is quadrant I and one is quadrant II. b) The given cosine ratio is positive. The domain –2π ≤ θ < 2π is two complete rotations. There will be four solutions, two in quadrant I and two in quadrant IV. c) The given tangent ratio is negative. This is true in quadrants II and IV. So, in the domain –360° ≤ θ ≤ 180° there will be three solutions, two in quadrant II and one is quadrant IV.

    θ

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 53 of 85

    d) The given secant ratio, and thus the cosine ratio, is positive. This is true in quadrants I and IV. In the domain –180° ≤ θ < 180° there will be two solutions, one in quadrant I and one in quadrant IV. Section 4.4 Page 211 Question 2

    a) For πθ3

    , the general solution is πθ 2π , where I3

    n n .

    b) For 5πθ3

    , the general solution is 5πθ 2π , where I3

    n n .

    Section 4.4 Page 211 Question 3 a) 2cosθ 3 0

    2cosθ 3

    3cosθ2πθ6

    Cosine is positive in quadrants I and IV. So, in the interval 0 ≤ θ < 2π, πθ6

    and

    11πθ6

    .

    b) csc θ is undefined when sin θ = 0. So, in the domain 0° ≤ θ < 360°, θ = 0° and θ = 180°. c) 2

    2

    5 tan θ 41 tan θ

    tanθ 1

    So, the reference angle for θ is 45°. In the domain –180° ≤ θ ≤ 360°, θ = –135°, –45°, 45°, 135°, 225°, and 315°. d) secθ 2 0

    secθ 21cosθ2

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 54 of 85

    The reference angle is π4

    .

    Cosine is negative in quadrants II and III.

    So, in the domain –π ≤ θ ≤ 3π2

    , θ = 3π4

    , 3π4

    , and 5π4

    .

    Section 4.4 Page 211 Question 4 a) tan θ = 4.36 θ = tan–1 4.36 θ ≈ 1.35 Tangent is positive in quadrants I and III. So, in the domain 0 ≤ θ < 2π, θ ≈ 1.35 and θ ≈ π + 1.33, or 4.49. b) cos θ = –0.19 θ = cos–1 (–0.19) θ ≈ 1.76 Cosine is negative in quadrants II and III. So, in the domain 0 ≤ θ < 2π, θ ≈ 1.76 and θ ≈ 2π – 1.76, or 4.52. c) sin θ = 0.91 θ = sin–1 0.91 θ ≈ 1.14 Sine is positive in quadrants I and II. So, in the domain 0 ≤ θ < 2π, θ ≈ 1.14 and θ ≈ π – 1.14, or 2.00 d) cot θ = 12.3

    θ = 1 1tan12.3

    θ ≈ 0.08 Cotangent and tangent are positive in quadrants I and III. So, in the domain 0 ≤ θ < 2π, θ ≈ 0.08 and θ ≈ π + 0.08, or 3.22. e) sec θ = 2.77

    θ = 1 1cos2.77

    θ ≈ 1.20 Secant and cosine are positive in quadrants I and IV. So, in the domain 0 ≤ θ < 2π, θ ≈ 1.20 and θ ≈ 2π – 1.20, or 5.08.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 55 of 85

    f) csc θ = –1.57

    θ = 1 1sin1.57

    θ ≈ –0.69 Cosecant and sine are negative in quadrants III and VI. So, in the domain 0 ≤ θ < 2π, θ ≈ π + 0.69 or 3.83 and θ ≈ 2π – 0.69 or 5.59 Section 4.4 Page 211 Question 5 a) 3 cos θ – 1 = 4 cos θ –1 = cos θ In the domain 0 ≤ θ < 2π, θ = π. b) 3 tanθ 1 0

    1tanθ3

    Tangent is negative in quadrants II and IV.

    In the domain –π ≤ θ ≤ 2π, θ = – π6

    , θ = 5π6

    , and θ = 11π6

    .

    c) 2 sin 1 0

    1sin2

    x

    x

    Sine is positive in quadrants I and II. In the domain –360° < θ ≤ 360°, θ = –315°, –225°, 45°, and 135°. d) 3 sin x – 5 = 5 sin x – 4 –1 = 2 sin x

    1sin2

    x

    Sine is negative in quadrants III and IV. In the domain –360° ≤ x < 180°, x = –150° and –30°. e) 3 cot x + 1 = 2 + 4 cot x –1 = cot x tan x = –1 Tangent is negative in quadrants II and IV. In the domain –180° < x < 360°, x = –45°, 135°, and 315°.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 56 of 85

    f) 3 secθ 2 02secθ33cosθ

    2

    Cosine is negative in quadrants II and III.

    In the domain –π ≤ θ ≤ 3π, θ = – 5π6

    , 5π6

    , 7π6

    , and 17π6

    .

    Section 4.4 Page 212 Question 6 Domain Interval a) −2π ≤ θ ≤ 2π θ 2π, 2π b) π

    3 ≤ θ ≤ 7π

    3 π 7πθ ,

    3 3

    c) 0° ≤ θ ≤ 270° θ 0 ,270 d) 0 ≤ θ < π θ [0, π) e) 0° < θ < 450° θ (0°, 450°) f) –2π < θ ≤ 4π θ (−2π, 4π] Section 4.4 Page 212 Question 7 a) 2 cos2 θ – 3 cos θ + 1 = 0 (2 cos θ – 1)(cos θ – 1) = 0 2 cos θ – 1 = 0 or cos θ – 1 = 0

    cos θ = 12

    or cos θ = 1

    In the domain 0 ≤ θ < 2π,

    θ = π 5π, 3 3

    or θ = 0

    b) tan2 θ – tan θ – 2 = 0 (tan θ – 2)(tan θ + 1) = 0 tan θ – 2 = 0 or tan θ + 1 = 0 tan θ = 2 or tan θ = –1 In the domain 0° ≤ θ < 360°, θ ≈ 63.435°, 243.435° or θ = 135°, 315°

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 57 of 85

    c) sin2 θ – sin θ = 0 sin θ (sin θ – 1) = 0 sin θ = 0 or sin θ – 1 = 0 sin θ = 1 In the domain θ [0, 2π),

    θ = 0, π or θ = π2

    d) sec2 θ – 2 sec θ – 3 = 0 (sec θ – 3)(sec θ + 1) = 0 sec θ – 3 = 0 or sec θ + 1 = 0 sec θ = 3 or sec θ = –1 In the domain θ [–180°, 180°), θ ≈ –70.529°, 70.529° or θ = –180° Section 4.4 Page 212 Question 8 Check for θ = 180°: Left Side = 5 cos2 θ Right Side = –4 cos θ = 5 cos2 180° = –4 cos 180° = 5(–1)2 = –4(–1) = 5 = 4 Left Side ≠ Right Side So, θ = 180° is not a solution. Check for θ = 270°: Left Side = 5 cos2 θ Right Side = –4 cos θ = 5 cos2 270° = –4 cos 270° = 5(0)2 = –4(0) = 0 = 0 Left Side = Right Side So, θ = 270° is a solution. Section 4.4 Page 212 Question 9 a) In step 1, they should not divide both sides by sin θ because this may eliminate one possible solution and if sin θ = 0 this division is not permissible. b) 2 sin2 θ = sin θ 2 sin2 θ – sin θ = 0

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 58 of 85

    sin θ (2 sin θ – 1) = 0 sin θ = 0 or 2 sin θ – 1 = 0

    sin θ = 12

    In the domain 0 < θ ≤ π,

    θ = π or θ = π6

    , 5 π6

    Section 4.4 Page 212 Question 10 sin θ = 0 when θ = 0, θ = π, or θ = 2π. However the interval (π, 2π) means π < θ < 2π. There are no values of θ for which sin θ = 0 in this interval. Section 4.4 Page 212 Question 11 The equation sin θ = 2 has no solution, because sin θ has a maximum value of 1. This is true for all values of θ, so the interval is irrelevant. Section 4.4 Page 212 Question 12

    The trigonometric equation cos θ = 12

    does have an infinite number of solutions. Cosine

    is positive in quadrants I and IV, so in one positive rotation θ = 60° and θ = 300°. However, all coterminal angles have the same value. In general, θ = 60° + 360°n or θ = 300° + 360°n, where n I. Section 4.4 Page 212 Question 13 a) Check by substituting θ = π into the original equation. Left Side = 3 sin2 θ – 2 sin θ = 3 sin2 π – 2 sin π = 3(0)2 – 2(0) = 0 = Right Side The solution θ = π is correct. b) 3 sin2 θ – 2 sin θ = 0 sin θ (3 sin θ – 2) = 0 sin θ = 0 or 3 sin θ – 2 = 0

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 59 of 85

    sin θ = 23

    In the interval θ [0, π], θ = 0, π or θ ≈ 0.7297, 2.4119 Section 4.4 Page 212 Question 14 Use n1 sin θ1 = n2 sin θ2 Solve for θ2 when θ1 = 35°, n1 = 1.000 29, and n2 = 1.33. 1.000 29 sin 35° = 1.33 sin θ2

    2

    12

    2

    1.000 29sin 35 sinθ1.33

    1.000 29sin 35θ sin1.33

    θ 25.56

    Section 4.4 Page 213 Question 15

    a) For sales of 8300, substitute y = 8.3 into π5.9 2.4sin ( 3)6

    y t

    .

    1

    π5.9 2.4sin ( 3)6

    8.3 5.9 πsin ( 3)2.4 6

    2.4 πsin ( 3)2.4 6

    π π ( 3)2 63 3

    3

    6

    8. t

    t

    t

    t

    tt

    Sales of 8300 air conditioners are expected in the sixth month, June. b) Graph the function

    π5.9 2.4sin ( 3)6

    y t

    .

    Minimum sales occur in the 12th month, December.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 60 of 85

    c) The formula seems reasonable. In western Canada you would expect sales of air conditioners to peak in summer and be least in winter. Section 4.4 Page 213 Question 16 9 sin2 θ + 12 sin θ + 4 = 0 (3 sin θ + 2)(3 sin θ + 2) = 0 3 sin θ + 2 = 0

    sin θ = 23

    θ = sin–1 23

    ≈ –41.810 314 9 The reference angle is 41.8°, to the nearest tenth of a degree. Sine is negative in quadrants III and IV. The solution in quadrant III is 180° + 41.8° = 221.8°. The solution in quadrant IV is 360° – 41.8° = 318.2°. Section 4.4 Page 213 Question 17 Examples: An equation such as cos x = 3 has no solution because the maximum value of cosine is 1. An equation such as sin x = –1, 0 < x < π, has no solution in the required interval, because

    sin–1 (–1) = 3π2

    .

    Section 4.4 Page 213 Question 18

    Given cot θ = 34

    , and θ is in quadrant III so x = –3 and y = –4. Using the Pythagorean

    theorem, r = 5.

    Then, sec θ = 53

    rx

    .

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 61 of 85

    Section 4.4 Page 213 Question 19 a) For the ball at sea level, substitute h = 0.

    π1.4sin3π1.4sin3

    π0 sin3

    π π πTherefore, 0, π, 2π, ...3

    0

    3 3

    th

    t

    t

    t t t

    So, in the first 10 s, the beach ball is at sea level at 0 s, 3 s, 6 s, and 9 s. b) The ball will reach its maximum height, for the first time, half way between 0s and 3s which is at 1.5 s. This will repeat every 6 s, so an expression for the time that the maximum occurs is 1.5 + 6n, n W.

    A graph of the function π1.4sin3ty

    confirms

    this reasoning.

    c) Since sine has minimum value –1, the minimum value of π1.4sin3ty

    is –1.4. So

    the most the ball goes below sea level is 1.4 m. Section 4.4 Page 213 Question 20 a) I = 4.3 sin 120πt Substitute I = 0, then 4.3 sin1 20π

    0 sin120πsinθ 0 at θ 0, π, 2π, ...0 120π 0

    1π 120π 120

    12π 120π 6

    0

    0

    tt

    t t

    t t

    t t

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 62 of 85

    Since the current must alternate from 0 to positive back to 0 and then negative back to 0,

    it will take 160

    s for one complete cycle or 60 cycles in one second.

    b) Each complete cycle takes 160

    s, so to reach the first maximum will take one-quarter

    of that time period or 1240

    s. As a decimal this is approximately 0.004 167 s.

    The cycle repeats every 160

    s, so in general

    the current reaches its maximum value at 1 1

    240 60n

    seconds, where n W. A graph

    of y = 4.3 sin 120πt confirms these results.

    c) The current is at its first minimum value at 3 1 1or 4 60 80

    s. As a decimal, this is

    0.0125. So in general, the current reaches it minimum value at (0.0125 + 160

    n) seconds,

    where n W. d) The maximum value of sine is 1, so the maximum value of this function is 4.3. The maximum current is 4.3 amps. Section 4.4 Page 214 Question 21

    π 3cos2 2

    x

    The reference angle for 1 3cos2

    is π6

    . Cosine is positive in quadrants I and IV.

    So, in the domain –π < x < π, π π π π or 2 6 2 6

    2π π or 3 3

    x x

    x x

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 63 of 85

    Section 4.4 Page 214 Question 22 a) The left side of the equation, sin2 θ + sin θ – 1, has the form x2 + x – 1 and cannot be factored.

    b) In the quadratic formula, x =2 4

    2b b ac

    a , substitute a = 1, b = 1, c = –1.

    sin θ 2( ) ( ) 4( )( )

    2(1 1 1 1

    1)

    1 52

    The only solution that makes sense is 1 52

    ; the other solution is less than –1 which is

    not a possible value for sine.

    c) From b), θ = 1 1 5sin2

    or θ ≈ 0.67.

    This is the solution in quadrant I. In the domain 0 < θ ≤ 2π, there is another solution, in quadrant II. θ ≈ π – 0.668 = 2.48. Section 4.4 Page 214 Question 23 a) The height of the trapezoid is 4 sin θ and the base of the trapezoid is 4 + 2 cos θ. Then, use the formula for the area of a trapezoid:

    sum of parallel sides2

    4 4 2(4cosθ) 4sin θ2

    (8 8cosθ)2sinθ16sin θ(1 cosθ)

    A h

    b) Substitute A = 12 3 in the formula from part a).

    h

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 64 of 85

    16sinθ(1 cosθ)

    12 3 sinθ(1 cosθ)163 3 sinθ(1 cosθ

    1

    )4

    3 3 sinθ(1 cos

    2 3

    θ)2 2

    Then, 3 1sin θ and cosθ2 2

    which is true when πθ3

    .

    c) Example: Graph the function y = 16 sin x(1 + cos x) and identify the first maximum. The graph shown here is in radian mode.

    Section 4.4 Page 214 Question C1 The methods and steps used to simplify linear and quadratic trigonometric equations are the same as those used for linear and quadratic equations. The major difference is the last steps when the inverse of trigonometric ratios have to be used and consideration of signs and domain is needed. Section 4.4 Page 214 Question C2 a) The point is on the unit circle if x2 + y2 = 1. For A: 0.384 615 384 62 + 0.923 076 923 12 = 1 Therefore A is on the unit circle. b) For a point on the unit circle, cos θ = x = 0.384 615 384 6 ≈ 0.385

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 65 of 85

    csc θ

    0.923 076 923 1

    1

    1

    1.083

    y

    0.923 076 923 10.384 615 384

    tanθ

    2. 006

    4

    yx

    c) θ = tan–1 (2.4) ≈ 67.4° This answer seems reasonable. The x-coordinate is about one-third the y-coordinate, so the angle is about two-thirds of the rotation from 0° to 90°.

    Section 4.4 Page 214 Question C3 a) Example: A non-permissible value is a value for which an expression is not defined. For a rational expression this is any value that would lead to division by 0. In the rational

    equation 5 9, 11

    xx

    .

    b) sin θtan θcosθ

    yx

    , so tan θ is not defined for values that make cos θ have value 0.

    c) In the interval 0 ≤ θ < 4π, cos θ = 0 when θ = π2

    , 3π2

    , 5π2

    , 7π2

    . These are the non-

    permissible values.

    d) In general, tan θ is not defined for θ = π π , I2

    n n .

    Section 4.4 Page 214 Question C4 a) 2 sin2 θ = 1 – sin θ 2 sin2 θ + sin θ – 1 = 0 (2 sin θ – 1)(sin θ + 1) = 0 2 sin θ – 1 = 0 or sin θ + 1 = 0

    sin θ = 12

    or sin θ = –1

    In the domain 0° ≤ θ < 360°, θ = 30°, 150° or θ = 270°.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 66 of 85

    b) The solutions are exact because both values for sin θ are special values. c) To check, substitute the solution into both sides of the original equation. Both sides should have the same value. Example θ = 270°: Left Side = 2 sin2 θ Right Side = 1 – sin θ = 2 sin2 270° = 1 – sin 270° = 2(–1)2 = 1 – (–1) = 2(1) = 1 + 1 = 2 = 2 Left Side = Right Side Chapter 4 Review Chapter 4 Review Page 215 Question 1 a) The terminal arm of 100° is in quadrant II. b) 500° – 360° = 140°; so the terminal arm of 500° is in quadrant II. c) 10 – 2π = 3.71, 3.71 – π = 0.57; so the terminal arm of 10 radians is in quadrant III.

    d) 29π 5π4π6 6

    ; so the terminal arm of the angle is in quadrant II.

    Chapter 4 Review Page 215 Question 2

    a) 5π 5(180 )2 2

    450

    b) 240° π240180

    4π3

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 67 of 85

    c) –405° π405180

    9π4

    d) 1803.5 3.5π

    630π

    Chapter 4 Review Page 215 Question 3

    a) 20° π20180

    0.35

    b) –185° π185180

    3.23

    c) 1801.75 1.75π

    100.27

    d) 5π 5(180 )12 12

    75

    Chapter 4 Review Page 215 Question 4 a) 6.75 – 2π ≈ 0.4668. The given angle is coterminal with 0.467 and terminates in quadrant I.

    b) 400° – 360° = 40°. The given angle is coterminal with 40° and terminates in quadrant I.

    c) –3 is almost –π. 2π – 3 ≈ 3.28 The given angle is coterminal with 3.28 and terminates in quadrant III.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 68 of 85

    d) –105° is a negative rotation, past –90°. 360° – 105° = 255° The given angle is coterminal with 255° and terminates in quadrant III. Chapter 4 Review Page 215 Question 5 a) All angles coterminal with 250° are given by the expression 250° ± (360°)n, where n is any natural number.

    b) All angles coterminal with 5π2

    are given by the expression 5π 2π2

    n , where n is any

    natural number. c) All angles coterminal with –300° are given by the expression –300° ± (360°)n, where n is any natural number. d) All angles coterminal with 6 radians are given by the expression 6 ± 2πn, where n is any natural number. Chapter 4 Review Page 215 Question 6 a) 80 000 rpm = 80 000(2π), or 160 000π radians per minute. b) 80 000 rpm = 80 000(360) degrees per minute

    80 000(360 )60

    480 000 /s

    Chapter 4 Review Page 215 Question 7

    a) 5π 3 1P ,6 2 2

    (1, 0)

    ,3 1

    2 2

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 69 of 85

    b) P(–150°) = 3 1,2 2

    c) 11π2

    is coterminal with π2

    , so 11πP2

    = (0, 1).

    d) P(45°) = 2 2,2 2

    e) P(120°) = 1 3,2 2

    f) 11π3

    is coterminal with 5π3

    .

    So, 11π 1 3P ,3 2 2

    .

    (1, 0)

    11π

    3

    (1, 0)

    ,1

    2 2

    3 120°

    (1, 0)

    2 2,

    2 2

    (1, 0)

    ,3 1

    2 2

    –150°

    ,1 3

    2 2

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 70 of 85

    Chapter 4 Review Page 216 Question 8

    a) The diagram shows πP3

    , with coordinates

    1 3,2 2

    . Then, 2πP3

    will be the same distance

    from each axis as πP3

    is, but in quadrant II. So, its

    coordinates are 1 3,2 2

    .

    4πP3

    will be the same distance from each axis, but in quadrant III. So, its coordinates

    are 1 3,2 2

    . Similarly, 5πP3

    will be the same distance from each axis, but in

    quadrant IV. So, its coordinates are 1 3,2 2

    .

    b) If 2 2 1P θ ,3 3

    , then θ is in quadrant II.

    πP θ2

    will be in quadrant III and will have coordinates 1 2 2,3 3

    .

    c) 5πP6

    is in quadrant II. Then, 5πP π6

    , which is a rotation of π more, will be in

    quadrant IV. 5πP θ P π6

    11πP6

    3 1,2 2

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 71 of 85

    Chapter 4 Review Page 216 Question 9 In the interval –2π ≤ θ < 2π:

    a) (0, 1) is on the y-axis, above the origin, so θ = 3π2

    and θ = π2

    .

    b) 3 1,2 2

    is in quadrant IV.

    θ = π6

    and θ = 11π6

    .

    c) 1 1,2 2

    is in quadrant II.

    θ = 5π4

    and θ = 3π4

    .

    d) 1 3,2 2

    is in quadrant II.

    θ = 4π3

    and θ = 2π3

    .

    Chapter 4 Review Page 216 Question 10 In the domain –180° < θ ≤ 360°:

    a) 3 1,2 2

    is in quadrant

    III. θ = –150° and θ = 210°.

    b) (–1, 0) is on the x-axis to the left of the origin. θ = 180°.

    (1, 0)

    ,3 1

    2 2

    (1, 0)

    ,1

    2 2

    3

    (1, 0)

    1 1,

    2 2

    (1, 0)

    ,3 1

    2 2

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 72 of 85

    c) 2 2,2 2

    is in quadrant II.

    θ = 135°.

    d) 1 3,2 2

    is in quadrant IV.

    θ = –60° and θ = 300°.

    Chapter 4 Review Page 216 Question 11

    a) Given P(θ) = 5 2,3 3

    , then θ is in quadrant IV.

    tanθ

    2 53 325

    yx

    Then, θ ≈ –42° or 318°, or (in radians) θ ≈ 5.55. b) θ terminates in quadrant IV.

    c) P(θ + π) will be in quadrant II with coordinates 5 2,3 3

    .

    d) P πθ2

    will be in quadrant I with coordinates 2 5,3 3

    .

    e) P πθ2

    will be in quadrant III with coordinates 2 5,3 3

    .

    (1, 0)

    2 2,

    2 2

    (1, 0)

    ,1

    2 2

    3

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 73 of 85

    Chapter 4 Review Page 216 Question 12

    If cos θ = 13

    , 0° ≤ θ ≤ 270°, then θ is in quadrant I.

    cos θ = xr

    , so x = 1 and r = 3.

    Determine y: x2 + y2 = r2 12 + y2 = 32 y2 = 8 y = 8 or 2 2

    Then, sin θ = 2 23

    yr tan θ = 2 2 2 2

    1yx

    csc θ = 3 2 or2 4

    32

    ry sec θ = 3

    13r

    x cot θ = 1 2 or

    42 2xy

    Chapter 4 Review Page 216 Question 13

    a) The terminal arm of 3π2

    is on the y-axis, above the origin. On the unit circle, its

    coordinates are (0, 1).

    sin

    11

    3π2

    1

    yr

    b) The terminal arm of 3π4

    is in

    quadrant II. Its reference angle is π4

    3πcos4

    12

    22

    2

    xr

    1

    2

    2

    2

    2 3π

    4

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 74 of 85

    c) The terminal arm of 7π6

    is in quadrant III. Its reference angle is π6

    .

    7πcot6

    3212

    3

    xy

    d) The terminal arm of –210° is in quadrant II. Its reference angle is 30°.

    sec (–210°)

    2

    13

    22 3 or

    33

    rx

    e) The terminal arm of 720° is coterminal with 0°. On the unit circle, its coordinates are (1, 0).

    tan 720° =

    10

    0

    yx

    f) The terminal arm of 300° is in quadrant IV. Its reference angle is 60°.

    csc 300°

    2

    13

    22 3 or

    33

    ry

    1

    1

    2

    3

    2–300°

    2

    3

    1

    2 30° 1

    6

    1

    2

    3

    1

    2

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 75 of 85

    Chapter 4 Review Page 216 Question 14 a) sin θ = 0.54, –2π < θ ≤ 2π. The sine value is positive, so θ is in quadrant I or II. θ ≈ 0.57, 2.57, –5.71, –3.71

    b) tan θ = 9.3, –180° ≤ θ < 360°. Since the tangent value is positive, θ is in quadrant I or III. θ ≈ 83.86°, 263.86°, –96.14°

    c) cos θ = –0.77, –π ≤ θ < π. The cosine value is negative, so θ is in quadrant II or III. θ ≈ 2.45, –2.45

    d) csc θ = 9.5, –270° < θ ≤ 90°. The cosecant value, and therefore the sine value, is positive, so θ is in quadrant I or II. θ ≈ 6.04°, –186.04°

    Chapter 4 Review Page 217 Question 15 a) sin 285° ≈ –0.966 b) cot 130° ≈ –0.839 c) cos 4.5 ≈ –0.211 d) sec 7.38 ≈ 2.191

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 76 of 85

    Chapter 4 Review Page 217 Question 16 a) Example: A(–3, 4) is in quadrant II. The angle of rotation measures approximately 125°.

    b)

    35

    cosθ

    or 0.6

    xr

    c) csc θ + tan θ

    15

    5 44

    ( 16)12

    112

    3

    r yy x

    d) θ = cos–1 (–0.6) = 126.9°, or 2.2

    Chapter 4 Review Page 217 Question 17 a) cos2 θ + cos θ = cos θ (cos θ + 1) b) sin2 θ – 3 sin θ – 4 = (sin θ – 4)(sin θ + 1) c) cot2 θ – 9 = (cot θ – 3)(cot θ + 3) d) 2 tan2 θ – 9 tan θ + 10 = (2 tan θ – 5)(tan θ – 2) Chapter 4 Review Page 217 Question 18 a) sin–1 2 is impossible because the sine value of an angle is never 2. Sine has a maximum value of 1.

  • MHR • 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 4 Page 77 of 85

    b) tan 90° is not defined because tan = yx

    and 90° is on the y-axis, so x is 0. Division by

    0 is not permissible. Chapter 4 Review Page 217 Question 19 a) 4 cos θ – 3 = 0

    cos θ = 3 or 0.754

    The cosine ratio is positive in quadrants I and IV, so there are two solutions in the domain 0° < θ ≤ 360°. b) sin θ + 0.9 = 0 sin θ = –0.9 The sine ratio is negative in quadrants III and IV, so there are two solutions (both negative) in the domain –π ≤ θ ≤ π. c) 0.5 tan θ – 1.5 = 0 tan θ = 3 The tangent ratio is positive in quadrants I and III, so there is one solution (in quadrant III) in the domain –180° ≤ θ ≤ 0°. d) csc θ is undefined. This occurs when y = 0. So, in the interval θ [–2π, 4π] there will be two angles which have their terminal arm on the y-axis in each of the three complete rotation, giving a total of six solutions. Chapter 4 Review Page 217 Question 20

    a) csc θ = 12, so sinθ2